Continuous Subarray Sum
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# LeetCode #523: Continuous Subarray Sum (C/C++).

medium

source: https://leetcode.com/problems/continuous-subarray-sum/
C/C++ Solution to LeetCode problem 523. Continuous Subarray Sum.

## Problem

Given an integer array nums and an integer k, return true if nums has a good subarray or false otherwise.

A good subarray is a subarray where:

• its length is at least two, and
• the sum of the elements of the subarray is a multiple of k.

Note that:

• A subarray is a contiguous part of the array.
• An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.

## Examples

### Example 1:

Input: nums = [23,2,4,6,7], k = 6
Output: true

Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.

### Example 2:

Input: nums = [23,2,6,4,7], k = 6
Output: true

Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.

### Example 3:

Input: nums = [23,2,6,4,7], k = 13
Output: false

## Constraints

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 109
• 0 <= sum(nums[i]) <= 231 - 1
• 1 <= k <= 231 - 1

## Solution

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 class Solution { public: bool checkSubarraySum(vector<int>& nums, int k) { int sum = 0; unordered_map<int, int> hashmods{ {0, 0} }; for (int i = 0; i < nums.size(); i+= 1) { sum += nums[i]; int mod = sum % k; if (!hashmods.count(mod)) hashmods[mod] = i + 1; else if (hashmods[mod] < i) return true; } return false; } };